1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Tension of a chain problem

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data

    A 20 m length of chain weighing 2.0 N/m is hung vertically from one end on a hook.
    Answer in Newtons

    1.What is the tension three quarters of the way up?

    2.What is the tension 1 m from the top?

    3.What is the tension 1 m from the bottom?

    2. Relevant equations

    F = ma

    w = mg

    3. The attempt at a solution

    Base on my imagine, it has 2 forces total. One is the weight of the chain which is mg (going down); and another one is Newton's law (going up) which is F = ma.

    But I look at the problem, it given mg = 2 N/m => m = 2 / 9.8 = 0.2 (what's units here??) I'm confused. Moreover, they give the length of the chain which is 20m. What can I do with it? The length??? I really dont know how to start!!!!
     
  2. jcsd
  3. Feb 23, 2010 #2

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The chain is not moving. What is its acceleration?? Then, Free Body Diagrams are a must. The basic weight unit is the newton, and the basic mass unit is the kg. Are you at all familiar with free body diagrams?
     
  4. Feb 23, 2010 #3
    The chain is not moving => the velocity is constant => you mean a = 0 m/s^2 ???

    I'm new at Physics so I'm still weak on it. I draw my diagram and came up with 2 forces.

    One is mg (pointing down) and one is N (pointing up). Then how can I do next?
     
  5. Feb 23, 2010 #4
    I suddently think of this:

    Given 2N/m, so we have length 20m => F = 2 N/m * 20 m = 40 N. Is this correct?
     
  6. Feb 23, 2010 #5

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That's the entire weight of the 20 m chain. If the chain weighs 40N, what force does the ceiling exert on the chain, in order that the sum of all forces acting on the chain are 0 (yes, it is not accelerating, or even moving at constant velocity; it just hanging there). Finding the ceiling force is always a good start before you examine the tensions at other points along the chain. You should note that if the chain weighs 2 N/m, then for example the first 10 feet of the chain weighs (2)(10) = 20 N, etc. Think Free Body Diagrams.
     
  7. Feb 23, 2010 #6
    If the chain weighs 40 N, the force is weight force = m * g
    I thought a Free Body Diagrams as:

    N - (m*g) = m*a (a = 0 ; m = 40N/9.8 = 4.08 kg)

    => 40N - (4.08kg * 9.8) = 0 N

    right????
     
  8. Feb 23, 2010 #7
    What is the tension three quarters of the way up?

    It means (40 N * 75)/ 100 = 30 N = answer for question 1. Correct?
     
  9. Feb 23, 2010 #8

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What I think you are trying to say is that

    N -mg = 0
    N - 40 = 0
    N =40 N, which is the ceiling force on the chain. This is a result of the application of Newton 1 to a free body diagram of the entire chain.
     
  10. Feb 23, 2010 #9

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    yes, correct!
     
  11. Feb 23, 2010 #10
    2.What is the tension 1 m from the top?
    It means 20m - 1m = 19m => (19m * 40 N)/20m = 38 N. Correct?
     
  12. Feb 23, 2010 #11

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes! I wonder what it is at that the very bottom (bonus question worth 10 points).
     
  13. Feb 23, 2010 #12
    I answers it but it's INCORRECT. I mean 38 N => incorrect somehow? What did I do wrong?

    The last question I did correct. It is 2 N/m * 1 m = 2 N.
     
  14. Feb 23, 2010 #13
    btw what do u mean "bonus question worth 10 points" ????
     
  15. Feb 23, 2010 #14

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Oh since you were on a roll with now all answers to the 3 questions corrrect, i just thought i'd challenge you with a 4th question I made up on my own, which if you answered correctly, and if I was your teacher, i'd give you extra credit. So my question to you is, if you choose to accept the challenge and answer it, is
    :surprised
     
  16. Feb 23, 2010 #15
    I think 40 N. Am I correct? LOL
     
  17. Feb 23, 2010 #16
  18. Feb 23, 2010 #17

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Sorry, no bonus. It's 40 N at the top, but 0 at the bottom. BTW, your answer of T= 38N at 1 m from the top is correct, i don't know why you were marked wrong;I'll give you a bonus anyway for being smarter than the book.....:wink:
     
  19. Feb 23, 2010 #18
    Got it. T = -38 N is more correctly. Agree?
     
  20. Feb 23, 2010 #19

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Oh, here we go again with the plus and minus sign.....tension forces always pull away from the objects on which they act....typically, tension Is considered a positive number...The tension at 1 m from the top is 38 n down acting on the top 1 m of chain; by newton's 3rd Law, the tension is 38 N acting up on the lower 19 m of chain at that same point. Just say T =38 N.
     
  21. Feb 23, 2010 #20
    But when I submit 38 N. It's INCORRECT! Why? I think because the last question is + 2 N. Agree?
     
  22. Feb 23, 2010 #21

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Now I know you've not been getting very good help with that question, but nonetheless, you shouldn't double post. I'll give it a look in the morning, under the original post, unless someone beats me to it....
     
  23. Feb 23, 2010 #22
  24. Feb 23, 2010 #23

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Oh, maybe, but the plus and minus sign really lacks meaning unless you identify the part of the chain you are looking at...give it a shot at neg 38 N....I'm maybe not as smart as a 5th grader :rolleyes:
     
  25. Feb 23, 2010 #24

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Gee, you're wearing me down...I've got to get up at 5....give it your best shot....19.6N is the max static friction available before it starts to slide..so what's F?? Good night , good morning, whatever...
     
  26. Feb 23, 2010 #25
    Hang on... I just submited -38, and INCORRECT. What's wrong??? I'm confused...
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook